Assume first that \(\kappa \geqslant 0\) . The space \(X\)  being \(\sigma \) -compact, there is an increasing sequence \((K_j)\)  of compact sets with the union \(X\) . The kernel being strictly pseudo-positive, for each \(j\)  large enough, we have \(0<c(K_j)<\infty \)  (cf. (REF ) and (REF )), and there is therefore an equilibrium measure \(\gamma _{K_j}\)  on the (compact) set \(K_j\)  (cf. footnote REF ). By the countable subadditivity of inner capacity on universally measurable sets [1]}, \(\kappa \gamma _{K_j}=1=\kappa \gamma _{K_{j+1}}\)  n.e. on \(K_j\) , hence \(\gamma _{K_j}\) -a.e. (Lemma REF ), and consequently \(\kappa \gamma _{K_j}\leqslant \kappa \gamma _{K_{j+1}}\)  on all of \(X\) , by the domination principle.
Thus the sequence \((\kappa \gamma _{K_j})\)  increases pointwise on \(X\)  to some function \(g\geqslant 0\) , and moreover
\(g=1\text{ \ n.e.\ on $X$},\) 
again by the countable subadditivity of inner capacity on universally measurable sets.
Given \(\mu ,\nu \in \mathcal {E}^+\)  with \(\kappa \mu \leqslant \kappa \nu \)  n.e. on \(X\) , the same inequality is fulfilled \(\gamma _{K_j}\) -a.e. (Lemma REF ); therefore, by Fubini's theorem,
\(\int \kappa \gamma _{K_j}\,d\mu =\int \kappa \mu \,d\gamma _{K_j}\leqslant \int \kappa \nu \,d\gamma _{K_j}=\int \kappa \gamma _{K_j}\,d\nu .\) 
By the monotone convergence theorem applied to the increasing sequence \((\kappa \gamma _{K_j})\)  of positive functions with the upper envelope \(g\)  (cf. (REF )), we obtain \(\int g\,d\mu \leqslant \int g\,d\nu \) ,
whence (REF ); for, \(g=1\)  \(\mu \) -a.e. as well as \(\nu \) -a.e. (Lemma REF ), \(X\)  being \(\sigma \) -compact.
It remains to consider the case of compact \(X\) ; then \(0<c(X)<\infty \) , and there is therefore an equilibrium measure \(\gamma _X\)  on the whole of \(X\) . Moreover, \(\kappa \gamma _X=1\)  holds true n.e. on \(X\) , hence \(\mu \) -a.e. as well as \(\nu \) -a.e. (Lemma REF ). Likewise, \(\kappa \mu \leqslant \kappa \nu \)  \(\gamma _X\) -a.e., the measure \(\gamma _X\)  being of finite energy. This altogether gives
\(\mu (X)=\int \kappa \gamma _X\,d\mu =\int \kappa \mu \,d\gamma _X\leqslant \int \kappa \nu \,d\gamma _X=\int \kappa \gamma _X\,d\nu =\nu (X)\) 
as desired.
